Question: Assume that $C$ is a positively oriented, piecewise smooth, simple, closed curve. Let $R$ be the region enclosed by $C$. Use the circulation form of Green's theorem to rewrite $ \oint_C 4x\ln(y) \, dx - 2 \, dy$ as a double integral. Choose 1 answer: Choose 1 answer: (Choice A) A $ \iint_R \dfrac{-4x}{y} \, dA$ (Choice B) B $ \iint_R 4x(y\ln(y) - y) \, dA$ (Choice C) C $ \iint_R 2x - 4x(y\ln(y) -y) \, dA$ (Choice D) D $ \iint_R 2x - \dfrac{4x}{y} \, dA$ (Choice E) E Green's theorem is not necessarily applicable.
Explanation: Assume we have a two-dimensional vector field $F(x, y) = P(x, y) \hat{\imath} + Q(x, y) \hat{\jmath}$ and a piecewise smooth, simple, closed curve $C$. Let $R$ be the region enclosed by $C$. Then the circulation form of Green's theorem states that we have the equality below: $ \oint_C P \, dx + Q \, dy = \iint_R \left( \dfrac{\partial Q}{\partial x} - \dfrac{\partial P}{\partial y} \right) dA$ Our first step should be to confirm that the given curve is compatible with using Green's theorem. The curve $C$ does satisfy all the conditions of Green's theorem. It is also positively oriented, which means we don't need to take the negative of our result at the end. The next step is to identify the components $P$ and $Q$ of the vector field $F$ over which we're integrating. We're simplifying from a line integral to a double integral, so we just need to match up the $dx$ term to $P$ and the $dy$ term to $Q$ : $\begin{aligned} & \oint_C 4x\ln(y) \, dx - 2 \, dy \\ \\ &P(x, y) = 4x\ln(y) \\ \\ &Q(x, y) = 2 \end{aligned}$ Our final step is to find $\dfrac{\partial Q}{\partial x} - \dfrac{\partial P}{\partial y}$. $\begin{aligned} &Q_x = 0 \\ \\ &P_y = \dfrac{4x}{y} \\ \\ &\dfrac{\partial Q}{\partial x} - \dfrac{\partial P}{\partial y} = \dfrac{-4x}{y} \end{aligned}$ Putting everything together, we get this double integral: $ \iint_R \dfrac{-4x}{y} \, dA$